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Why can you only hear an aircraft after the shockwave passes?
Why do you only hear engines and other noise from a jet going supersonic after the plane has passed? A lot of resources use this picture to illustrate :

This is cool and all, but say you changed the geometry of the nose of the aircraft in a way (presumably changing the flow turning angle) that made the shock at a sharper angle. That would make it so you heard the aircraft later after it passed compared to the one with the standard nose geometry, despite the same speed of both aircraft. (I think)
Basically I'm asking why you only hear the plane after the Mach cone passes.
Or you could also do a theoretical scenario, where an infinitely thin plate (at 0 AoA) was going supersonic through the air, and it had something like a speaker on it. When would you hear the speaker after the plate had passed? There is no shockwave so you wouldn't hear that.
((I understand that sound won't travel upstream when going supersonic, but I'm curious as to why the shock forms at the exact point where you can hear sound from said object))
Some of the sources I was looking at :

If the plane accelerates to the speed of sound or faster, then the sound waves can no longer move fast enough to get out of the way of the aircraft. The waves bunch up and combine to form a shock wave. This then trails behind the aircraft in a V-shape. Something similar is seen for the boat on water.

from here. (this one didn't seem to descriptive, but it provided an interesting way to look at it)
Also, the Wikipedia article on shockwaves has some interesting points

Downvoter, what on this page did I not follow correctly?

I didn't downvote this, but there is a certain amount of repetitive similarity in your questions.  Have you ever seen a boat wake, or been rocked by the waves?  Do the waves hit you after the boat has passed, or when is it directly abeam you?  Think about it... this one seems quite obvious.

@MichaelHall yeah good points, but what still confuses me is if you change the shape of the nose of an aircraft, that will change the shock angle, presumably changing when you hear the plane. All of this at the same speed, which is the really confusing part.

"if you change the shape of the nose of an aircraft, that will change the shock angle" -- not true, the angle depends entirely on the speed of sound and the speed of the aircraft.

You sure? Maybe read this answer to see where I'm coming from.
(I still might be wrong)

I am speaking of the apex angle of the supersonic shock cone, as shown in the third figure of your illustration. This is what determines when you hear the sound of a supersonic aircraft passing overhead. -- The effects of the oblique shock angle do not extend more than a few yards from the supersonic object.

@A.I.Breveleri hmm okay. That’s a little confusing, because of this answer. Especially the first 2 paragraphs. Wouldn’t that make shocks straight?

@A.I.Breveleri Basically the flow turns to make a shockwave from an object. Then that turned flow turns the flow above it, continuing the process. Then that goes on, so I’m not sure how it’d change within the first few meters of the aircraft.

Also perhaps this picture of a T-38 will show you what I mean. It looks like the effects of the oblique shock angle extend for a bit more than a few yards. Am I missing your point?

The shockwave itself is just noise from the engine. If the Mach cone hasn't hit you yet, the sound has not had enough time to travel from the source to the observer. The question seems to be asking why you can't hear a sound wave before it reaches you?

Why do you only hear engines and other noise from a jet going supersonic after the plane has passed?

If the object making noise is far enough, then you always hear it after it has passed, no matter if it is super or subsonic.
Looking for example at the picture in your question, an observer on the ground would hear the airplane only when the biggest "sound bubble" reaches it and this always happens when the airplane has already travelled forward a bit in the meantime, it's just a geometrical matter and it has nothing to do with being supersonic. So, no matter the speed, an observer on the ground will always hear the airplane with a delay, delay which is obviously bigger the faster the airplane is.

Anyway I agree that some of your sources are definitely misleading since they call the cone on the right part of your picture "shock wave". When the sound bubbles coalesce together behind a supersonic airplane they form a "Mach cone", not a shockwave. A shockwave is a completely different phenomenon which depends, among other things, also on the shape of the aircraft. The angle of the Mach cone depends instead only on the speed of the airplane according to the following equation $\alpha=arcsin(1/Ma)$ (source Wikipedia):

You seem to have such a grasp on this question that it seems impossible that you don't actually understand.  I'd really like to understand what is missing between what you know and what you're trying to understand.
The sound of the aircraft (from engines, whatever) travels at the speed of sound.  This is also the fundamental sound that all pressure disturbances travel at.  The speed of sound is through the medium -- i.e. through the air.  Which means the sound travels the same speed no matter what creates it (or how fast the noise source moves).
A supersonic aircraft (one that flies faster than sound) will fly faster than the sound it makes.
Imagine that the aircraft was 5000 ft (about a mile) away from you -- but headed directly at you.  The noise it makes right now will take about 5 seconds to reach you (speed of sound is about 1000 ft/s).  If the aircraft is traveling at Mach 1.5, then it will take 3.3 seconds for the aircraft to reach you.
Of course, the aircraft is constantly making noise -- but it is traveling faster than all of that noise.
If the aircraft is pointed directly at you -- such that it crashes into you -- then at zero ft. of separation, it would make noise -- and the noise and the aircraft would hit you at the same instant.
Of course, real aircraft are flying overhead (not directly at us).  So, we're listening to the sound as it propagates down to us on the ground.  We also don't have as good of a perception of when the aircraft is directly overhead.

I see. The part I am a little confused about is why the flow turning (that changes with speed) and therefore determines the shock angle, determines when you’d hear the aircraft. Couldn’t you have a shock whose angle is such that you hear the shock before you hear the sound? Like there would be a delay from when you heard the shock and the sound of the aircraft.

A shock can not possibly form at that angle.  Look at the oblique shock chart https://en.wikipedia.org/wiki/Oblique_shock#/media/File:ObliqueShockAngleRelation.png  The greatest a shock angle can be is 90 degrees -- which is a normal shock.   If that shock propagated to the ground, you would hear it at the same time as the disturbance passed.  However, normal shocks cause zero turning (but the downstream flow is subsonic).  Because of all this, normal shocks usually only exist near the body and are curved to form oblique shocks away from a body.

"Couldn’t you have a shock whose angle is such that you hear the shock before you hear the sound?" You always hear the shock before you hear the sound... because the shock is the sound.

Ah I see. What else confuses me is the reason why changing the nose cone shape wouldn't change the shock angle. As said here, the oblique shock angle is determined by the Mach number and the amount of turning needed. Wouldn't the amount of turning needed change with a different nose cone, and therefore change the shock angle?

Yes, absolutely.  Look at the oblique shock chart again.  Changing the shape of the nose will change the shock angle.

Yeah okay that’s what I was thinking. The part of that that is confusing is that you can change when you hear an aircraft without changing the speed. All you have to do is change the nose cone shape.

This is the way I am looking at it: The nose cone shape will determine the shock angle, also determining when you hear the aircraft. So if the aircraft is traveling forward supersonic, the sound waves can't go forward, they have to go back. Speed is the only thing that I think changes this. If that's the case, why does changing the nose cone shape, (which changes shock angle), affect when you hear the aircraft? Why isn't the shock heard separate from the aircraft itself?

The shock is heard separate from the aircraft itself.  The shock travels at the speed of the aircraft.  Lets say the aircraft was flying low enough that you would hear both the sonic boom and the engines.  If the aircraft is flying very subsonically, you would hear the engines before the aircraft passes (and there would be no boom).  If the aircraft is flying supersonically, then the aircraft would pass first, you would then hear the boom (because you are off-track some distance), and then some time later you would hear the engines.

Ah okay. So if the shock angle was increased (more straight), would that increase the time that it took for you to hear the engines after the shock had passed?

Assuming changing the angle was done by changing nose cone geometry

I think I might've answered my own question. There was a video I was watching of an F-15 going around Mach 1.1 or so. When it was approaching the camera, I was able to hear the engines, then the shockwave. There were no other jets around either. Is this proving you can hear the sound before the shockwave?

No, there must have been some other sound.  The aircraft is traveling faster than sound, you could not hear the engines before the shockwave.

Responding to the more straight shock question -- very slightly, but only slightly.

Ah okay. I think I understand it more now. So because sound from the engines and aircraft can’t travel through the shockwave, you only hear the engines after the shock?

if you change the shape of the nose of the aircraft ...

It seems that is what NASA is doing here.
So now one is reducing shock wave amplitude (how loud it is), not the shape of the cone, which depends on Mach number.
This becomes more obvious when one changes the frame of reference to an observer on the ground.  Since sound will propagate at, well, the speed of sound, the shape of the cone depends entirely on the aircraft speed.
One may reason this by geometry from the fixed point of the observer a fixed distance away from the aircraft when it passes.
Where will the aircraft be at Mach 1.2 compared with Mach 2.4 when it's "sound" reaches you?  The angle from the observer to the plane is the angle of the shock wave.
This concept is not restricted to supersonic flight.  The change of frequency of sound waves emanating from a passing object is known as the Doppler effect.

curious why the shock wave forms at the exact point where you can hear the sound.

The shock wave is the sound.  Our eardrums perceive  pressure change of the wave as noise.

when would one hear the noise of the speaker after the plate had passed?

Well it depends what your carrier medium is.  Vibrations made in air are restricted to the speed of sound just as the propagation of the whine of the jet engine is.  But radio or light waves travel much faster.  Thankfully, vision and radar enable us to see and communicate ahead even at speeds greater than Mach 1.

Fun fact:  people flying at Mach 2 in the Concorde can speak to each other fine.  But their voices cannot carry ahead of the aircraft in flight.  A bit of a mind bender, but not unlike a glider with a "rate of descent" less than the upward vertical wind speed in a thermal.

@Wyatt based on your recent questions: could the sound (drag) "barrier" be partly the result of shock waves causing turbulence in the boundary layer of the aircraft?

I think I see what you mean. So the amount of flow turning needed is less, (for the X-59) but the shock angle stays the same?

"curious why the shock wave forms at the exact point where you can hear the sound."
"The shock wave is the sound."  Upvoting for this clear, concise, (and should be obvious...) statement.  It isn't just a coincidence Wyatt!

@Wyatt I read the article about flow turning.  This may be a local phenomenon applied to (supersonic) air intakes.  Any sound propagated from any angle cannot be faster than the lead shockwave, regardless of shape.  I worry that the Quesst designers are in for a nasty surprise when they scale up to airliner size as lift creation (larger wings) may create a much bigger boom.


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Will the air be low pressure above an airfoil going less than Mach 0.3?
If flow only becomes compressible (>5%) above Mach 0.3, and the only way for air to accelerate is by having a pressure gradient¹, will the air be low pressure on top of a wing going less than Mach 0.3?
So to explain that better, say you have a plane traveling Mach 0.2. The flow isn't compressible, hence, no air is being compressed. The way an airfoil works, (as I know) is when the air contacts the front portion of the airfoil, it compresses and therefore accelerates along the airfoil turning the air low pressure.
If the flow is not able to compress, then how does the air accelerate? The only way for air to accelerate is an adverse pressure gradient¹, which isn't possible below Mach 0.3.

¹: I'm not 100% that this is true, however I see no other way for air to accelerate. Seeing that I'm not sure, I might be mistaken.
Also, having a ‘pressure gradient’ might not be the correct wording. A pressure difference between 2 connected packages of air might be more accurate.

"when the air contacts the front portion of the airfoil, it compresses and therefore accelerates along the airfoil turning the air low pressure." That's not how it works. And you can have a change in pressure without compressibility. This happens in columns of water all the time.

@DavidK When the flow reaches the front of the airfoil, what else would it do if it didn’t accelerate? The air might not have a static pressure that is above ambient, but there does have to be a high pressure initially to accelerate the air, no?

There just needs to be a pressure differential. For example, ambient pressure far in front of the wing and a lower pressure above the wing.

@DavidK Right, but what is going to make that low pressure? Or how is that low pressure going to be made?

How does air accelerate?  Push it with higher pressure or pull it with lower pressure (or both).

That is not the way an airfoil works. To move through the air, you have to push the air out of the way. Newtons 3rd law says anything that gets pushed, pushes back. As you know, gravity is pulling you down all the time, but you have to push the air below you out of the way to actually start falling. If you, or your "airfoil" (or a brick) pushes that air down at least as hard as gravity, then you don't actually fall and that is called lift.

Understand that airflow is always compressible. The 0.3 Mach number is just a dividing line where compressibility effects are no longer a significant term. This allows for teaching simplification and what we all know as non-compressible flow. Make sure you have that concept clear. In the real-world if you are asked to go design X with a max velocity of 0.25 M -- you will be expected to consider all terms.

flow only becomes compressible above Mach 0.3... Say you have a plane traveling Mach 0.2. The flow isn't compressible... If the flow is not able to compress, then how does the air accelerate?

There's a mixing up of terminology at work here.
Pressure $p$ and density $\rho$ are two different physical quantities. Being a fluid incompressible only means that the changes in its density $\rho$ (and not pressure $p$) are negligible but this is not directly related to the changes of pressure around it - and viceversa. "Inviscid" might be a better wording.
The ubiquitous Bernoulli equation, for example, is strictly valid only for incompressible fluids:
$p+½\rho V^2=constant $
And it is more than enough to calculate the aerodynamic loads around a subsonic aircraft. If comprimibilty is no more negligible, then the compressible form of the Bernoulli equation is used:
$h+½V^2=constant $
where $h$ is the enthalpy.

I see. So even below Mach 0.3, the flow still has pressure changes big enough to make an effect?

@Wyatt: Insects fly

Well, we have a chicken and egg issue.
We know by Reynolds number analysis that there is indeed lower pressure above the wing where V is less than Mach 0.3.
Typically at Re = around 500,000 one has an increase in L/D ratio attributed to "top lift".
For a 1.5 meter wing, this would be around 5 meters/second or around 12 miles per hour!
So what accelerates the air?  The same thing that lifts the plane!

lower pressure created above the wing, above the Reynolds threshold, begins to accelerate the air

Lower than this, lift is only created by air deflecting off the bottom of the wing.  For a fixed geometry wing, the threshold is determined by velocity.
Air impacting the front of the wing at Mach < 0.3 is deflected away, not compressed.  The wing passes by the air molecules as they try to fill the void (lower pressure) above and behind (Angle of attack) the wing.
The air molecules accelerate downwards, but (ideally) never reach the wing (it has already gone by).  The result is downwash aft of the wing and an upforce (lift) on the wing.

source: Stack Exchange

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What are the Boeing 777 aural alerts indicating in this ATC recording?
There was a recent incident (avherald.com) with an Air France Boeing 777 on approach to CDG where several different aural warnings can be heard in the ATC audio recording (YouTube) when the pilot has the mic keyed. They went around and ultimately landed safely, but I'm just curious what all of the different aural warnings that can be heard in the recording in the linked video indicate.
Note: I am specifically not asking for speculation about the cause(s) of the incident. That is explicitly off-topic here. I'm only asking what the different aural alerts that can be heard are designed to indicate.

A video about the incident made by a 777 pilot. He comments on the audio recording from about 8 minute mark. https://youtu.be/cslSQB5mgyc (also explains the procedures and mentions a couple of possible reasons for the problem)

I can only hear 2 distinct warning sounds in the video:

There is a high-low tone audible from 0:31 to 0:35 (also heard later). This is the master warning siren.
There is a low-to-high sound audible from 0:48 to 0:50 (also again around 0:57). This is the autopilot disconnect wailer.

My 777 FCOM groups both of these sounds in the same category (under siren):

Aural
Light
Calls Attention To

Siren
Master WARNING lights
EICAS Warning Message:

AUTOPILOT DISC

CONFIG DOORS

CONFIG FLAPS

CONFIG GEAR STEERING

CONFIG PARKING BRAKE

CONFIG RUDDER

CONFIG SPOILERS

CONFIG STABILIZER

OVERSPEED

CAIN ALTITUDE

PILOT RESPONSE

STABILIZER

Beeper
Master CAUTION lights
New EICAS caution message, except:

ENG SHUTDOWN

ENG SHUTDOWN L

ENG SHUTDOWN R

(Boeing 777 FCOMv2 15.20.5 - Warning Systems - System Description)
There are however company options available to play the wailer instead of the siren for autopilot disconnect, meaning some airlines get different aural alerts.
I cannot hear the master caution beeper in the video.
References for all sounds (YouTube links are timestamped):

Master Warning Siren: YouTube (0:46 - 0:49), example of PILOT RESPONSE in-flight (thanks @ymb1): YouTube (43:21 - 43:30)
Autpilot Disconnect Wailer: YouTube (0:07 - 0:09)
Master Caution Beeper: YouTube (0:04 - 0:05)

The third sound I was referring to is a high-pitched chirp that I can hear at least a couple of times in the background behind the pilot breathing in between the Master Warning and the AP disconnect. If you start listening around 0:50, there's what sounds like a chirp right around 0:54 and again at 0:56 before the AP disconnect sounds again.

@reirab OK, I just listened to that part again. I think I know what chirp you mean, but I'm not familiar with it as an aural alert in the 777. It might be something partially cut-off by the transmission or even an artifact of poor AM radio quality. Not sure though...

It sounds to me like the beeper which indicates a Caution message on the EICAS. That could be any of a large number of conditions.
A Warning message from EICAS is accompanied by a siren, and it didn't sound like one to me, but even that could be any of a number of conditions.

If you play the video at 50% you will hear the beeper as well as a hi-lo tone. I thought at first it was the master caution warning, but I don't have the 777 manuals with me to confirm.

@Raffles There seem to be at least 3 different tone patterns in the video. There's at least a high-low, a high-pitched chirp, and a repeated low-to-high ramp-up (latter is AP disconnect, maybe?)

The Caution beeper sounds like this.

The Hi-Lo tone is the Auto Pilot disconnect aural warning. I can't identify the other coming later in the vidéo.

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Will the 3D effect change shockwave angle/strength?

Similar to how the 3D relieving effect will change the shockwave angle (see above), will this same effect change the shock angle and strength further away from the body? As a shock goes through different temperature air and density, the shockwave strength will be affected. Will the 3D effect also change the angle/strength?
Imagine you have a cone going supersonic like the one above. As the shock gets further from the body, it will continue in a straight line. However, shocks aren’t just 2d, they continue out in a cone shape.
Similar to how if a sphere was emitting sound in all directions, the sound would be forced to spread out as it got farther from the sphere. I’m asking if this same effect happens to shockwaves, considering shocks are in a 3d cone shape.
(If the question is hard to understand, comment and I’ll do my best to improve it; I found this hard to explain.)

Think of the cone not in isolation but as a combination of the cone itself plus the deadwater it carries along. I don't know if that helps because I found it hard to isolate the real question here.

@PeterKämpf I don’t blame you, I wasn’t able to explain this very well. Imagine you have a 2d circle. It is emitting sound waves in all directions. In this case for the sake of explanation, when the sound waves travel off, they go in a straight line and don’t interact at all. Because each of those waves were emitted at a slightly different angle (because it was a circle) at some distance wouldn’t those waves start to separate? I’m asking if that same thing applies to the (oblique) shocks that extend from a body.

It's a single, circular wave which is emitted. Amplitude will drop with distance, but other than that it will stay unchanged as it radiates out. Since it is radially symmetric, I do not understand why you think something should separate here.

@PeterKämpf Right, but each part of that single wave was emitted at a slightly different angle. Imagine you were standing on a part of the theoretical circle, and  you throw a ball straight up. Now move over a tiny bit and do the same thing (still on the circle). Those balls would be on trajectories leading away from each other, right? Sorry if this is still confusing, I can't explain this for the life of me.

The imaginary rays of sound start to "separate" instantly as they propagate, not "at some distance". There is not a break or gap in the expanding circular/spherical wave front because there are an infinite number of rays and between every two rays that are separating, there are an infinite number of rays that fill in the gaps. Still, the density of the rays (which are just mathematical models - there aren't real sound rays) decreases as the wave front expands and that decreasing density models how the intensity of the wave front decreases as it expands.

Is the cone an infinite cone?  If the cone is finite -- like all real cones -- then there will ben an expansion fan off of the end of the cone that will interact with the shock and will bend it.
Remember, the shock at an angle turns the flow by a certain angle.  If the flow does not need to turn the same amount, then the angle of the shock will change.

Hmm, okay. I’ll copy/paste what I said to Mr. Kämpf here, because I didn’t explain my question properly. Imagine you have a 2d circle. It is emitting sound waves in all directions. In this case for the sake of explanation, when the sound waves travel off, they go in a straight line and don’t interact with each other at all. Because each of those waves were emitted at a slightly different angle (because it was a circle) at some distance wouldn’t those waves start to separate from each other? I’m asking if that same thing applies to the (oblique) shocks that extend from a body.

(Let me know if that didn’t make sense)

The circular pressure wave that comes from a disturbance does not come at an angle, it goes in all directions equally.  The reason they can coalesce into a Mach wave is because the disturbance is moving.  It is the combination of the forward speed with the radial wave speed that causes the waves to build up in front of the disturbance and spread out behind.  Again, this is for an infinitesimal disturbance -- not for a cylinder or sphere.

It sounds like the thrust of your question is about whether the wave front will begin to develop 'gaps' like in your analogy of the ball being thrown up from different angles. The issue is that the wave front is not a discrete object like the balls, but rather the force moving through the objects (which in this case are air molecules). If you were talking about photons from the sun, then you may see gaps between the individual photons that radiate out in a sphere as you move very far out, but because the shock-wave is transmitted through the movement of atmosphere, without significant displacement of the molecules that are transmitting the force, the shock-wave will behave more like a continuous line in space with infinitely many points all spreading out. As such, we would not perceive any spacing between the points, just a net depreciation of amplitude as you move further from the origin of the shock-wave.
In the case of oblique shocks, I am not sure if you could have constructive/destructive interference from multiple incidence points of the geometry of the object that is moving. That may appear to be gaps, but I think they would also behave more like lines/regions in 3d space - growing in scale as they propagate away from the origin.
I hope I haven't misunderstood what you were getting at.


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Is it possible to tell what aircraft this is?
On a flight a few years ago, I got this picture. Is it possible to tell the aircraft by just this image? I’ve found a few similar but they are a bit different than this one. Someone I was with said they saw “Airbus” on the headrests of the seats, but I’m not sure because I didn’t.

When trying to find it myself, it kind of looks like the wing of the A320 or A321. But then it also looks like the wing of some 737’s.
I know that there are some planes that use the same wing as others, so this might not be possible to tell.

Additional information:
The airline was JetBlue (relevant because some airlines don’t operate certain planes)
It had 2 seats on each side of the cabin
The flight lasted around 1 hour (Different planes are used for longer or shorter haul flights)

My guess: A type of 737.

If you still have any documentation from the flight (e.g. the e-mail sent out when you purchased the ticket), you can probably confirm exactly what type of plane it was.  Ticket stub or boarding pass would also have it, but people are less likely to hold onto those for that long.

Can't be a 737, those have today a 3+3 seat configuration, not 2+2 (unless maybe some VIP versions?)

@kebs yeah you’re correct. I was just going off of the wing geometry, but I should’ve seen the seat config as well.

It's an EMB-190. Jetblue only flies two aircraft with two seats on each side of an aisle, the A220 and the EMB190.
Looking the wing underside the EMB-190 has three flap track fairings and one engine pylon fairing visible at the wing trailing edge.

The A220 has only the three flap track fairings visible, the engine pylon ends midway back under the wing and is not visible.

According to this source, Jet Blue operates a fleet of A220, A320, A321, ERJ190.
Do you remember the seating arrangement in coach?  Most likely a 3-3 (3 seats on either side of a single aisle) arrangement, but if it was a 2-1, 3-2, or a 2-2, that would be significant.
From looking at some of the photos, it looks to me like an ERJ-190.
This photo shows the paint job on the upper surface of the wing.
The other most similar contender is the A321-200, but the leading edge of the winglets on those is left un-painted.

Yeah it had 2 seats on each side of the cabin. I think it is in fact a 190, considering the other answer. Thanks.


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Will a plane 'shake' when it stalls?
I haven't been able to confirm this, but does a plane shake when it stalls? If so, why?
I was able to find some info on this, but from a non-confirmed source.

As the plane shakes and rattles, it's a telltale sign that something is awry with the airflow over the wings.

From here.

I've also heard of this happening when the plane is maneuvering / pulling hard. (for example a fighter jet in a dogfight)
Does this actually happen, and if so, why?
(My assumption is that it's due to the turbulence over the wings when maneuvering or in a stall)

... and this turbulence hitting the tail, yes.

@PeterKämpf That would definitely make sense, thanks.

Will a plane 'shake' when it stalls?

Yes - for the simple reason that it has been designed to do so.
When designing an airplane (at least of conventional shape i.e. with a fuselage, a wing and a tailplane) the wing is laid out so that at high AoA the root stalls first while the center sections and its tips continue to produce lift.
This is beneficial in two ways: the stalled wake living the root impinges on the tailplane making it shake and warning the pilot about the imminent stall. Plus the ailerons and the other movable surfaces located on the outer part of the wing keep on working and the airplane remains maneuverable.

Riding onboard the 747-400 during stall testing, for example, I can assure you - it shook! The short answer upfront is the unsteady nature of separated flow, especially when coupled with a (relatively) flexible wing. Think of how you are buffeted when driving behind a large truck on the highway. Obviously, the frequency and amplitude of buffeting will vary with aircraft, loading and flight condition. I've heard a distinguished fighter test pilot say repeatable buffet characteristics are a good cue regarding angle-of-attack and energy state - "tactically useful" he said.

So let’s look at your question briefly from the standpoint of a high performance propeller driven aircraft.  Herein is shown the V-n diagram for this aircraft across the flight envelope for velocity $V$ and wing loading $n$ from sea level to altitude (40k ft).  The source is Dommasch, et. al, 1967.  This flight envelope is dynamic and expands to its maximum structural and velocity limits at sea level, or shrinks to its more limited range of flight conditions at altitude.  The notation defining n literally means the wing loading in $g$ units.  Hence, this is also called a V-g diagram.
Among the salient features to the question at hand are the limiting conditions noted for $n = (V/V_s)^2$ shown for maneuvers incurring both positive and negative wing loadings.  These are the stall limits.  Velocities shown in this diagram indicate true speeds in mph, and under the stall-limiting conditions shown in the diagram, would indicate stalling speeds.  For instance, an attempt at sea level to perform a maneuver at 200 mph which will sustain a wing loading of 5g will result in a stall.  Under such conditions, think turbulent separation and consequent buffeting resulting from the stall.  Further, a sea-level attempt to incur a loading exceeding 8g at 300 mph will result in damage to, or failure of the aircraft structure. The element of greatest concern, in this case, is the wing.  Now think about the consequences of incurring a stall under these dynamic and exceedingly harsh flight conditions in a lower-elevation, high-density atmosphere. Wing flexure and stall buffeting can be extreme.  Notice also that at higher velocities, for flight conditions that are well away from the stall-limiting velocities, wing loadings can be incurred that exceed the structural limit.
Notice now at higher velocities the indicated buffet limit due to the presence of Mach shocks on the wings.  These buffeting conditions are incurred across the range of flight conditions at the limiting velocity of the aircraft.  Notice also that favorable for the pilot at altitude, the aircraft can develop its full range of flight performance without incurring wing loadings and stresses that would result in structural failure.  But also notice that under these conditions, the aircraft will be pressed to its limits both in exceeding limiting lift coefficients and g loadings thereby resulting in stall, or otherwise in limiting velocities resulting in Mach buffeting.  Such conditions, of course, can be seen to become more hazardous as altitude decreases toward sea level (as indicated above).


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What is the purpose of thinner airfoils used near wing tips?
I know that a thicker airfoil at root will strengthen the root spar but what is the reason behind using thinner airfoils at the tip?
Is it because thinner airfoils have higher stalling angles of attack? If so, why?
Comparing both thick and thin airfoils, who generates more lift and why?
Also, my final question is that in what kind of designs would such a change of airfoils be recommended? I'm currently working on designing a high lift wing (selig s1223 airfoil) for delivering a payload of about 15kgs. with the total aircraft weighing about 20-22kgs. Would changing the airfoils in my wing design be recommended?

Thicker airfoils delay stalls - not thinner. The thicker the leading edge radius the higher angle of attack before stall.

There are two main reasons for thinning the wing towards the tips:

Thinner airfoil sections cause less drag simply because they disturb airflow less.
Thinner sections down to 10% have a higher maximum lift coefficient, which helps both near stall and during maneuvering: A section with a higher maximum lift coefficient will show more aileron effectiveness, too.

You hit it -- the thinner airfoil has the same effect as the wing twist known as "wash-out", that of delaying the stall at the tip.  This is generally desired, because a tip stall tends to initiate a spin, and a spin at low altitude has a high fatality rate (if only for the airplane, in the example you give), where a staight-ahead stall leaves some slight chance to recover.

It is much better and more effective to use more area at the tips in order to delay stall there. The best designs add tip sweep, too.


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Do high-wingers have dihedral?
I know that high wingers have an inherent dihedral effect.
Are there any examples of high wing aircrafts with a dihedral?
If so, what is the reason behind it?

This is not a well-researched question-- a look at some 3-view diagrams or a walk down a GA airport flight line would have given the answer.

High wing planes typically have less dihedral than an otherwise comparable low wing design, because of keel effect: they have more righting moment for a given dihedral angle because the fuselage hanging below the wing increases pressure near the wing root on the low side when a bank induces slip.
The value of the dihedral can be low enough that the wing may look flat (Piper Cubs and Taylorcraft look flat, to me), but if you build a high wing airplane with genuinely zero dihedral, the wing will appear to droop (take a look at period photos of Lindbergh's Spirit of Saint Louis).
Dihedral is present in nearly all high wing aircraft, for the same reason it's present in nearly all low wingers -- because it tends to support or raise the low wing when a bank without coordinated rudder causes side slip.  This produces lateral stability, which in turn makes the airplane easier to fly.

Depends mainly on the shape and size of the fuselage, and if the plane has wing sweep or not.

The high wing aeroplane has a wing/fuselage interference effect that tends to roll the aeroplane away from the sideslip direction, the low wing aeroplane wants to roll into the sideslip direction.

The second contributor to roll moment from side slip angle is from swept wings. The figures above are from my treasured copy of prof. Gerlach's uni book, the second one shows the difference in air speed over the wings, and therefore the difference in lift causing a stabilising rolling moment, away from the side slip.
Dihedral (positive V-shape) provides a positive roll-sideslip stability, required on low wing aeroplanes. With high wingers, at larger fuselage diameters the stabilising effect from interference and wing sweep can be excessively high, and then anhedral must be applied.
Image source
Other high-wingers do benefit from dihedral.
Image source

please add that wing sweep adds a rolling moment with sideslip of its own, which mostly explains the anhedral of the An-225, so I can upvote this answer.

Yes, look at any Cessna 150, 172, 177, 180. All have dihedral. It helps the plane return to a wings level state after a disturbance.

Yes, many if not most high wing aircraft (general aviation at least) have dihedral. The reason is the same as for low wing aircraft--to improve stability. I had a remote controlled high wing slope soarer glider that I built which had 0 dihedral. It was very maneuverable, wings tended to stay banked, unlike other aircraft with more dihedral that have stronger tendency to level out.
The dihedral on a high wing aircraft, be it positive, none, or negative (as illustrated below) will depend on the engineering and performance goals of the aircraft being designed. If you're building a primary trainer its going to have a lot of dihedral, even if its a high wing. If you want maneuverability then it will have less. In the case of the Harrier below, I imagine there were performance and engineering reasons for the extreme anhedral.
Here is a high wing airplane (Harrier), and it actually has a negative dihedral (i.e. anhedral).

Wikipedia

Yep, this and two or three other combat jets (F-104, etc.), plus one or two military transports (C-5A?).  The question as I read it wasn't, however, whether high wing designs always have dihedral, but do they ever have it, and why?

Good point, updated.


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Why are cambered airfoils used for horizontal stabilizers in some designs?
From what I know, usually NACA9/12 or similar symmetrical airfoils are used for the tail.
Still some aircrafts use highly cambered airfoils for elevators.
What are some advantages and disadvantages of this configuration?

It's done to get a little more lift coefficient, and hence more tail downforce power, at low speeds and there is a drag penalty to take depending on how much camber.  The Zenith STOL family are the extreme example, and being draggy flying lunchboxes anyway, the penalty is minor in relation to the benefit and allows a smaller, lighter tail unit than would be needed otherwise.

Some high performance and transport aircraft have cambered airfoils for the tail but the amount of camber is subtle (a bit more curvature on bottom than top) and you can only tell if you can see the actual airfoil profile, or examine the surface up close.

John K, do you have an example of some cambered airfoils used in horizontal stabs?

The CRJ/Challenger jets have cambered tail airfoils.   I can't find a picture where it is obvious just looking at it though.

Apologies, I wasn't meaning to challenge your answer! To the contrary, I was hoping you would know of a specific airfoil so I can plug it into my CAD. Try to theoretically do better than a symmetric airfoil when we know that the tail will generally be providing negative lift.

For the same reasons as camber is used is the main wing: to extend the maximum lift coefficient and the stall angle of attack. The following figure is from Abbott & Von Doenhoff, and shows the pressure distributions over both a cambered and an un-cambered wing profile: higher pressure differential = more lift.


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How to calculate the amount of weight my plane can takeoff with?
This is how I approached this problem-
My takeoff run is limited. So, depending on my configuration I calculated the maximum possible velocity I can attain in my groundroll. I assumed that at the end of the run my lift should be greater than the weight I want to carry. So I computed the lift I generate at the end of the groundroll. I got the CL value from xfoil for my angle of incidence. Would this method give me a rough estimate of the total weight my plane can carry?

What does the POH for your plane say it is for the atmospheric conditions you are taking off in for a short field takeoff?

Yes, you can do your equations at OEM (Operating Empty Mass) and get your ground distance. Then you can calculate the same for different mass values (your OEM+your payload), and generate a distance vs mass. Since you know how much distance you have (including any safety factors), you can see what mass you can take off with. Or work backwards from the same equation, set the maximum distance you allow, and work back to your mass value.
Keep in mind this will not tell you whether the airplane is structurally strong enough to carry that weight, only if it can take-off with it. Structural calculations are a bit more complex than this. Also, you might want to do similar calcs for landing distane as well!

How did you calculate the maximum velocity achieved in ground roll, without knowing weight on board?  It seems unlikely that your approach is going to lead to a realistic estimate of how much weight you can safely get off the ground from a short runway-- I think you are going to need to find some performance tables based on real-world data, or create your own.
IF your calculation of max velocity at end of ground roll was somehow correct, then your approach MIGHT lead to a very conservative estimate of how much weight you can safely lift.  The reason for this is that you appear to be basing your calculation on the fuselage being in a level attitude, not nose-high relative to the flight path -- (you said "for my angle of incidence".)  In other words you appear NOT to be basing your calculations on the MAXIMUM lift coefficient that your airfoil can achieve.  But not knowing all the details of your calculations, it would be impossible for anyone to really give a reassuring answer to your question.

I suppose you probably calculated maximum velocity for several different values of weight, or as a function of weight-- still that doesn't seem like an easy thing to do correctly as many different variables are involved


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Why does a turbulent boundary layer cause more friction drag than a laminar boundary layer?
From what I understand, friction is caused by viscosity and a bigger Reynolds number signifies lower viscosity.
Thus, a higher Reynolds number(i.e. turbulent flow) results in lower friction.
But, turbulent boundary layer is said to have a much steeper speed gradient at the aircraft skin, causing much more friction drag.
These two statements seem contradictory. What am I missing here?

Possible duplicate of this question:
(https://aviation.stackexchange.com/questions/36678/how-does-reynolds-number-affect-skin-friction-drag?rq=1 )

No, a higher Reynolds number only signifies a lower ratio of viscous to inertial forces. Since you increase the Reynolds number by either increasing speed or length, both will drive up the inertial forces, making the friction forces relatively lower.
Let's do a Gedankenexperiment: Consider an infinitely thin flat plate which moves through air in its lengthwise direction, so there is no pressure gradient along the flow path. The friction force caused by the boundary layer is F. Now add some length at the trailing edge of the plate. The added surface will add some more drag F', so the total drag will be F + F', both being positive. Since the length has increased, the Reynolds number of the flow has increased in the same way, but so has the friction force.
What has decreased is the coefficient of friction, however, because the thicker boundary layer towards the end of the flat plate will cause less friction force per unit of length. The coefficient is calculated by dividing the force by the dynamic pressure and the surface area, and since the surface area has increased by more than the drag, the coefficient of friction is lower.
You can repeat the experiment with a flow speed increase, and now the friction force will go up, too. So both ways to increase the Reynolds number will increase friction drag. Again, the drag increase is lower relative to the increase in dynamic pressure because the boundary layer at the higher speed is thinner, and again only the coefficient of friction is lower.
Only now, when that is out of the way, I want to answer your question directly: If you compare fully turbulent boundary layers, an increase in the Reynolds number will lower the coefficient of drag but will increase friction drag as outlined above.
Now compare two similar fully laminar flows at two different Reynolds numbers, and again the flow with the higher Reynolds number will experience more friction drag. The different speed profiles of their respective boundary layers will only be significant when you compare a laminar flow with a turbulent flow at the same or similar Reynolds number. In the range of Reynolds numbers typical for aircraft the turbulent flow will cause more friction drag.
If you now repeat the initial Gedankenexperiment with a laminar boundary layer and add some length at a point where the local Reynolds number is safely above 500,000, the local flow will already have transitioned to turbulent and the added length will add a fully turbulent portion. Now it can happen that the added surface will experience more drag per unit of area than the forward, mostly laminar surface. The result is both a higher friction drag and a higher coefficient of friction.
If you allow me to copy the illustration from this answer here, the effect can be easily seen:

Friction drag coefficient over chord for an E502mod airfoil at 3° AoA. Blue: Top surface, Red: Bottom surface.
Now this plot uses a real airfoil with a changing pressure coefficient over length, but with a flat plate the plot would look similar. The much higher local speed gradient at the surface of the boundary layer in the rear part will add much more drag per length and the total drag will increase disproportionately. However, a drag decrease with increasing Reynolds numbers is nowhere to be seen.

Good question.
The thing is in the velocity profiles for laminar and turbulent boundary layers. Lets look at the picture below. The profiles are little different. The turbulent profile is "fatter", or fuller, than the laminar profile. For the turbulent profile, from the outer edge to a point near the surface, the velocity remains reasonably close to the freestream velocity,then it rapidly decreases to zero at the surface. In contrast, the laminar velocity profile gradually decreases to zero from the outer edge to the surface.

The key now is in the wall shear stress. More shear stress results in higher skin friction drag. The wall shear stress is defined as a product of viscosity coefficient [mu] and the velocity gradient at the wall:
$$ \left.\frac{dV}{dy}\right|_{y=0} $$
Which is the reciprocal of the slope of the curves at the surface. It is clear that the velocity gradient near the surface for laminar flow is smaller than for the turbulent one, thus wall shear stress for the laminar flow is smaller than for the turbulent one.
This means that laminar flow has smaller skin friction drag than turbulent flow due to faster velocities near the surface.

You merely repeat what Zaber already alludes to in the question: "But, turbulent boundary layer is said to have a much steeper speed gradient at the aircraft skin, causing much more friction drag."

The basic reason is that the energy to create the turbulence comes in the first place from the forward motion of the aircraft or air vehicle.


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Which airfoil would be optimum for a glider?
The glider I am designing has a mission requirement to fly for as much time as possible. It will fly in the low Re range of about 150,000 to 250,000.
While comparing different airfoils I am looking for a high value of Cl^3/2/Cd to get as high endurance as possible and a high Cl/Cd ratio.
Any other parameters I should consider in selecting the airfoil?
I am currently considering Wortmann FX 63-137, E61, DAE-31 and Fx-74-c15-140.
Any other airfoil suggestions would be appreciated.

How do those airfoils compare to other record setting airfoils?  For example, https://en.wikipedia.org/wiki/Thomas_Knauff  World glider records
Single-place glider Out-and-return distance: 1,646.68 kilometres (1,023.20 mi), 25 April 1983, Schempp-Hirth Nimbus 3
Distance over a triangular course: 1,362.68 kilometres (846.73 mi), 2 May 1986, Schempp-Hirth Nimbus 3
Free distance using up to 3 turn points: 1,394.04 kilometres (866.22 mi), 1 June 1993, Schempp-Hirth Discus B

For human-carrying gliders, the task hasn't been duration since at least the 1940s -- tasks are cross-country speed over a defined course.

Related, maybe a dupe?

For a particular task of optimizing an airplane design for a specific point, you will need to consider the following characteristics, together with the endurance parameter:

The pitching moment of the airfoil, (which will define the tail design, hence will affect weight and CG range and controllability)
Tolerance to contamination and geometric impurities (especially CFD based methods may converge to non-robust shapes)
Stall behavior (does it stall mildly or violently, as a wing)
Wing weight (spar location, spar web thickness)
Aeroelasticity (wing structural modes, static and dynamic aeroelastic properties)
Manufacturability (if somebody will really build it!)

Many of these would require working on a conceptual or preliminary sizing task, where you would design a planform, twist, dihedral, tail volume, CG, mass estimation, aeroelasticity computations etc.
Number 1-2-3 could be more vital and would help pick an airfoil with less efforts than the others.


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What is the relation between hinge moment at the control surface and servo torque?
I am trying to estimate the required servo torque for my design using hinge moment at the elevator.
XFOIL results were as follows-

I am getting a Hinge moment of about 0.06-0.08 Nm. for the given deflection.
How can I relate this to the servo torque I require? Will they be equal?

Its a perfectly valid question with a good and to-the-point answer. Hinge moments are part of flight control design.

The servo torque depends on the mechanism. If it has a gear ratio of 1, they will be equal. If the servo is driving larger angular deflections than its own, the servo torque will be higher.
Another important point is to consider the dynamic forces. XFOIL only computes results for steady conditions. Dynamically, the servo may have to bear even larger torques; due to acceleration or gravity.
Finally, unless you really trust your computations, some uncertainty factor can be used to reduce the risks of redesigning the control system solution.


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What is the string antenna on the back of this Cessna 172 for?
Okay, so this question was asked before. But I didn't get the answer I wanted. I read that its some sort of antenna. what is it, what does it do, and how does it work? Also, is it in anyway related to the antenna on the to of the 707s vertical stabilizer?

The one on the 707 is probably the HF antenna... https://aviation.stackexchange.com/questions/24688/what-is-this-wire-going-from-mid-fuselage-to-the-tail-on-this-737

I was almost sure the one on the Cessna was HF as well, but it may be a loop antenna, might need a better picture of it.

https://aviation.stackexchange.com/questions/15506/what-radio-frequency-bands-are-associated-with-the-long-wire-antennas-why-do-so

So I read the article. It seems to be an adf antenna. But I saw a photo of a pc12 with one. What would be the purpose of that?


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What information does a pilot have about aircraft operating on an intersecting runway?
This incident at O'Hare from July 2006 made me wonder what information is available to a pilot (specifically, in an airliner) regarding aircraft that are using an intersecting runway.
In this incident, the controller had a United 737 line up and wait on 27L because the United flight wasn't ready to go yet while an Atlas 747 was on final for 14R. Once the United pilots radioed that they were ready, tower cleared them for takeoff on 27L, despite the fact that the Atlas 747 was touching down on 14R and was about to cross 27L on the roll-out. The Atlas flight did call out that he was on final, but that's the only indication I can see from the video that the United pilot has regarding the position of the Atlas flight. Judging from the below airport diagram, it appears that the ORD terminal would have been between the two aircraft at the time the United pilot was cleared for takeoff, so the aircraft wouldn't have been able to see each other until they approached the intersection.

Obviously, the controller shouldn't have cleared the United flight for departure until Atlas was out of the way. However, given that he received the takeoff clearance from Tower, I'm wondering whether the United pilot would have had any other tools at his disposal to detect the position of the Atlas 747 and realize that the controller made a mistake (before he saw a giant 747 roll out in front of him on the runway and before the controller realized his mistake.) Specifically, do airline pilots have any additional means for determining the position of other aircraft on intersecting runways aside from relying on ATC, listening to the Tower frequency, and looking out the window?
I'm aware of TCAS for use in-flight, but I'm not familiar enough with it to know if it would help in an intersecting runway situation like this where both aircraft are on the ground.

As the question mentions, good situational awareness is an excellent safety tool. ATC is responsible for runway separation at towered airports, but pilots can play an important part of backup in the case of a human mistake.
The United pilots may or may not have been able to see the Atlas 747 on TCAS, as the terminal building that obstructed their view could also have blocked the transponder signal. TIS-B (Traffic Information Systems - Broadcast), a part of ADS-B, provides traffic both in the air and on the ground. ADS-B was not available in 1996, but if the same situation happened today and the United aircraft had an ADS-B signal on the surface and displayed TIS-B information, they would see the Atlas 747 on their displays. Whether or not that information would be useful to avoid an incursion in a cluttered runway environment is unclear.
The FAA is implementing a system to provide exactly the type of warning you're describing. Runway Status Lights (RWSL) use surface RADAR and lights embedded in the pavement to detect conflicts and alert pilots that a runway is unsafe to enter or take off from. In this situation, the United pilots would have seen red takeoff hold lights to inform them that the runway is unsafe for departure. The FAA plans to have RWSL operational in 17 airports, including Chicago O'Hare, by 2017.

TIS-B = Traffic Information Service Broadcast. It is not a system in the aircraft. The United 737 does not have ADS-B IN, so they did not have a display of ADS-B or TIS-B traffic.

Your linked FAA article says The FAA also plans to deploy Runway Status Lights at the following airports by 2011 : Atlanta, Baltimore Washington International, Boston, Charlotte, Chicago O’Hare, Denver, Detroit, Washington Dulles, Fort Lauderdale, Houston Intercontinental, New York John F. Kennedy and La Guardia, Las Vegas, Minneapolis, Newark, Orlando, Philadelphia, Phoenix and Seattle So it should already be in place at O'Hare

Updated with a better link about RWSL. I had a news release from 2009; the FAA "re-scoped" the project in 2013, and on the RWSL page ("last modified August 11, 2014"), ORD is still "Scheduled to receive RWSL".

Specifically, do airline pilots have any additional means for
  determining the position of other aircraft on intersecting runways
  aside from relying on ATC, listening to the Tower frequency, and
  looking out the window?

Basically the answer is no. The only thing that could have been of help is TCAS, however:

Aircraft on the ground are not visible on TCAS; the Atlas 747's TCAS would be unable to see the United 737 on the ground.
TCAS works line of sight, if the Atlas 747 was obscured by the terminal there is high probability that the aircraft was not visible to the United 737's TCAS system. If it was, the range and direction measurements were unreliable.

Some airports have Advance Surface Movement Guidance & Control Systems (A-SMGCS) that indicate whether runways and taxiways are safe to use. However only the highest level (A-SMGCS level 4: Conflict Resolution, Automatic Planning & Guidance) includes automatic guidance for the pilots, lower levels only provide guidance to controllers.
In the future ADS-B will be of help. In 2020 all commercial aircraft in the US will be equipped with ADS-B OUT. To display the traffic information in the cockpit, aircraft need to be equipped with ADS-B IN as well. As it stands, equipping with ADS-B IN is voluntary. Just like TCAS, ADS-B works line off sight. However, airports are likely to install re-transmitters at strategic points to overcome signal obstruction by buildings.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


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Does the pressure recovery on the rear side of an airfoil recover all of the pressure?
I was wondering if all of the low pressure created by the airfoil gets recovered on the rear side of the wing. (I know during a stall or flow separation, this doesn’t happen, I think)
My main question is if you put a pressure sensor right behind the trailing edge of an airfoil, would the pressure be the same as ambient, higher, lower? How about if you put a pressure sensor in the wake of the wing, what would that say? Does AoA or speed affect anything?

I put it in a comment because it doesn't answer all of your question. No, it doesn't recover completely because of the boundary layer that doesn't allow the airflow to close at the trailing edge.

[Does] all of the low pressure created by the airfoil gets recovered on the rear side of the wing?

Only in inviscid flow. Then we speak of a rear stagnation point.
In reality, viscous effects cause enough energy loss that the static pressure at the trailing edge rises only a bit above its ambient value from the confluence of upper and lower flow and then quickly drops to ambient pressure when you follow the streamline from the trailing edge into the wake. See below for the XFOIL pressure coefficient plot of the HQ-17 airfoil.

If you put a pressure sensor right behind the trailing edge of an airfoil, would the pressure be the same as ambient, higher, lower?

That depends on the direction in which the sensor is pointing. Normally, those sensors are placed in flow direction so they measure the total pressure (sum of static and dynamic pressure). A rake of those is used to measure the pressure distribution over the height of the wake which is a very good way to determine the drag of airfoils. In the wake, total pressure is quite a bit below its freestream value due to the slowing down of flow speed in the boundary layer.
If you point the sensor in a direction orthogonal to flow speed, it should measure static pressure. With attached flow the pressure would be slightly higher than ambient and with separated flow it would drop below ambient pressure.

thanks for your answer.     “In the wake, total pressure is quite a bit below its freestream value due to the slowing down of flow speed in the boundary layer.”
I thought the boundary layer was just a slowed down layer that experiences the effect of friction? Wouldn’t the pressure in the boundary layer still recover on the rear side of the airfoil? (My assumption was that the pressure in the boundary layer was higher than the outer flow, because of it not moving as fast, so higher pressure)

@Wyatt there is a speed dent in the wake which gets broader and shallower downstream until it dissipates completely. A bit of the airplane speed continuously gets transferred into a big chunk of air.

I see. A little late to comment, but coming back to this answer I had a small question. So I understand how high pressure is made when the upper and lower surface flows confluence at the trailing edge, but how is that related to viscous effects? I thought viscous effects just means the friction between the fluid layers. In other words, how would inviscid flow make the high pressure not happen?

@Wyatt Inviscid flow produces the rear stagnation point, so it does "make the high pressure happen". With viscous effects the energy in the flow is reduced so pressure recovery remains incomplete.

Ah okay. So viscous effects means the friction between the layers of the fluid. Pressure recovery relates to the pressure recovering over the surface of a wing. So how would the viscosity of the fluid change how the pressure recovers over the wing? Even if the energy is reduced, why would that change how the pressure recovers? (I might end up asking this as a separate question as it might not be answerable in a comments section) ((Also this will be my last question; I don’t want to ask tons of questions))

@Wyatt Remember what happens when flow separates. Here the deceleration in the boundary layer is less severe, so the air still flows downstream, but slower than ambient. At the trailing edge this remaining speed produces only a minor pressure maximum, and total pressure in the wake behind the wing will show a bucket where the ex-boundary layer is.


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Where are the vortices from the trailing edge on the wing of a plane?
I always knew planes made these vortices when going through clouds/smoke, but I thought they were just tip vortices. Looking further into it, they are way too big to be tip vortices, so they had to be from the wing (trailing edge). I thought that the vortex from the wing was positioned perpendicular to the trailing edge of the wing, the same way tip vortices are perpendicular to the tip of a wing, if that makes sense. In that case, the vortex would not be parallel to the trailing edge of the wing, like in this photo. (Try to put some of it into layman's terms if you can, not a professional, thanks!)
(last thing, I've heard they rotate counter-clockwise, why is that?)

Clockwise or counter clockwise depends on whether you are looking at the right wing or the left wing, and looking from the front or the back.

@RTO Yes, that helps a lot, thanks! Still am wondering though, do the vortices I was talking about still exist? Like how the tip vortices are perpendicular to the tip of the wing, wouldn't that work for the trailing edge also? (still assuming the answer you gave me is true from the downwash)

@RTO Ah okay, sounds good.

Your intuition is right: the trailing edge releases vortices both perpendicularly and parallel to it. The formers are proportional to the spanwise change of lift (the tip-vortices are the most famous but there's much more) while the latter are proportional to the change of lift in time. If it helps I could elaborate a proper answer.

@sophit oh okay makes sense. If you could put some of this into the answer section so I can go back and look later on, I’d really appreciate it.

The vortex you're thinking about is what we call the 'startup vortex'  It is shed when the circulation changes -- the largest such change is when you go from rest to impulsively starting a lifting low.
For the most part, we ignore the starting vortex as it is infinitely far away during a cruising steady flight.  In an inviscid ideal world, it exists forever.  In the real world it dissipates and fades away.

ah okay probably also bigger at higher aoa I’m assuming?


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Is this "separation" caused by the boundary layer?
I was wondering if the void seen in this picture (on top of airfoil, not the wake from the airfoil) is from the boundary layer thickening, or from the fact that as you go further back on an airfoil, the pressure rises, "bending" less air towards the wing. Which causes it and why?
(It might be neither of those ideas, I'm no professional)
Thanks!

Do you have any idea about the Reynolds Number at which that picture was taken? For that low an angle of attack there should be no separation at the usual speeds.

Unfortunately no, but here’s the link to the video it was from : https://m.youtube.com/watch?v=rCpZpKZLz14

@PeterKämpf (it’s at timestamp 2:00 in the video)

This answer should also help.

It is both due to the boundary layer and the not perfectly spaced smoke trails injected in the airflow: if you watch carefully, there's an additional trail underneath the airfoil which is missing above it.

Ah I see, thanks. Happy thanksgiving!

The nomenclature of the art is laminar --> boundary layer --> turbulence and flow separation
As we can see in the video, the first smoke line above does bend less towards the trailing edge, but maintains its form until past the trailing edge.
So, yes, the boundary layer will thicken, and there is drag$^2$, but this is a consequence of creating lift.  Heavier turbulence eventually breaks the pressure gradient.  Ideally, you want this to happen after the wing has passed.
This early NACA video$^1$ seemed to focus on ways to reduce drag for higher speed flight, and is worth watching in its entirety.
$^1$ see 1st comment by Wyatt below question
$^2$ higher aspect ratio wings eliminate some of this less productive area, resulting in a higher Lift to Drag ratio at the expense of some strength.

Ah yeah I did notice that the flow line above didn’t go turbulent until the wing had passed. Thanks, and happy thanksgiving!


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Do wake vortices interfere with the flow over/under the wing?
So as the title suggests, I was wondering if the wake vortices formed by a plane interfere with the flow over/under a wing. (My assumption is that they don't, but I'm not sure)
Also, how would speed and angle of attack affect this?
Picture to show what I mean by wake vortices

Wake vortices start over and under the wing.
Air moves towards low pressure and away from high pressure. A lift-creating, straight wing has its lowest pressure right in the middle while below it pressure is relatively high. Therefore, the streamlines converge towards the middle on the upper and away from it on the lower side. When that air leaves the wing at the trailing edge, it has an inward component above and an outward component below it. All what is needed for the vortices to form is to be free of that separating layer called "wing".
A different and less intuitive explanation uses bound vortices to explain lift. Since those vortices have no end, but the wing has, they leave the wing in backward direction (and are then called free vortices). At every point where this vortex strength changes, bound vortices bend backwards and leave the wing. This vortex sheet then combines into the two vortices visualized by water droplets in the picture of your question at some distance behind the wing.
Both explanations are equivalent and mean the same. Lift creation over a limited wingspan will cause those vortices, and the pressure distribution over the wing (which produces lift) produces those vortices as an inseparable byproduct as well. To call this "interference" would be inappropriate.
Vortex strength is inversely proportional to the square of flight speed and proportional to the square of span loading (mass per unit of wingspan). When doubling the lift by doubling angle of attack (counted from the zero-lift angle), vortex strength will quadruple.

I see, thanks for your answer. One question I have though: if there is an outward component of flow on the bottom of the wing and an inward component on the top, that would create a vortex along the wing, right? Like a vortex along the span somewhere, not the tip vortex.

@Wyatt No, since the overwhelming speed is still backwards. The streamlines will only bend slightly and gain a small sideways component over the wing. Only after they leave the wing does the full rotation develop.

ah, I see, thanks. I was curious about how swept wings would fair with the airflow bending inward (on the top of the wing) due to them having a spanwise outward flow component already. Would they just mathematically add, meaning the spanwise flow would be a little less due to the inward bending air?
Also, I don't want to ask tons of questions, but how come the pressure is lowest in the middle of a wing?

@Wyatt Good question about swept wings! Short answer: Poorly! The center of swept wings has much less lift per area than the inner wing because streamlines cannot bend inwards there ("Mitteneffekt"). Streamlines bending outward is a different, boundary-layer-related phenomenon. Why lowest pressure in the middle of a straight wing? Because there the distance to the tips is greatest there and flow is more 2D than farther out.

That makes sense, thanks! I think I had misinterpreted what you had said about the lowest pressure in the middle of a wing, but now that makes total sense. Thanks again Peter, you always answer my questions!

(Generalisations/simplifications do apply here...)
They do - or to be more precise are more of less an effect of an interfered airflow on the wing - but it depends where on the wingspan.
Vortexes are more or less created by a (partially) spanwise flow of air near the wingtips to equalize the pressure diffential below and above the wing.
See https://en.wikipedia.org/wiki/File:Tip_vortex_rollup.png for a graphic on that subject.

So yes, there is a section on the wing where the airflow is disturbed by the generation of the vortices.

"Vortexes are more or less created by a (partially) spanwise flow of air near the wingtips" - more precise: They are shed by the wing wherever the local lift changes over span. This effect is strongest at the tips, but happens over the whole span.

@tsg oh okay thanks. So basically the vortices I’m talking about pull some air spanwise? How would that happen because those vortices are formed below the wing, correct?

To be clear because I think there possibly could be some confusion: I’m talking about the vortices in the wake of the plane, not the tip vortices.

The  question doesn't really make sense in terms of real physics. There is only one real flow field created by the passing of the plane. There aren't multiple features that can interfere with with each other via action at a distance.
When modelling flow fields with potential flow models, we do use imaginary vortex points or lines or sheets; or imaginary dipole points or lines or sheets. And these do affect the flow over the entire domain - action at a distance. But real flows don't have any action at a distance.
To get the best results using potential models, some of these imaginary entities have to be sprinkled in the wake to get the wake to develop properly. In this sense, within the potential model, wakes can interfere with the flow over the wing. But this is basically a kludge or exploit of the model to get better agreement with real behavior.
The real vortices are inextricably linked to the production of lift; and their creation requires energy that manifests as a drag component of the plane. If you try to actively mess with the vortices in an attempt to reduce its associated drag component, you can't - the energy has already been spent. For instance, installing a counter-rotating prop to try to counter the vortex with propwash swirl doesn't reduce drag.


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Twist optimization for a low speed aircraft
I'm currently designing an aircraft with the following specifications-
s=1.8m^2, AR=7-8, trapezoidal planform with a taper ratio of about 0.45, a load of 25KGs and a cruising speed of 13-15m/s.
I'm barely getting any improvement in drag reduction or tipstall delay after testing twists upto -4deg.
A twist of -3 is delaying tipstall by about 0.7deg while reducing Cl by 0.08 in xflr.
Is a twist even necessary for my model?
Also I've read that a twist optimizes drag reduction and lift distribution for a single aoa in non-elliptical planforms.
Should this angle be the aoa for the Cl I get for my cruise speed?
Also how would my planes angle of incidence depend on the twist that I select?

Is a twist even necessary for my model?

Yes, with that amount of taper I would recommend twist. Those 4° look rather low on first sight. The tapered wing will produce induced loads on the outer wing due to the amount of suction of the deep inner wing.  Below is a plot of where flow separation starts for different sweep angles and aspect ratios (source):

I would classify your wing as a high taper wing (second from top, right column). Without twist it will start to stall from the tip. My recommendation is to use 0.7 taper even if that means higher root bending moments. With 0.7 taper and the same stall angle of the inner and outer wing airfoil, 2° of twist are usually sufficient.

Twist optimizes drag reduction and lift distribution for a single aoa in non-elliptical planforms. Should this angle be the aoa for the Cl I get for my cruise speed?

Yes.

Also how would my planes angle of incidence depend on the twist that I select?

Proportionally to the lift loss on the outer wing with increasing twist, the whole wing needs more angle of attack for the same lift. You will, however, need many degrees of twist before the wing's incidence will increase by more than one degree.


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How do endplates reduce induced drag?
I have heard the explanation which states that the endplates reduce the tip vortex formation thereby reducing induced drag. From what I understand induced drag is the backwards component of the total force acting on the wing and a consequence of lift creation over a finite span. Thus it depends on the downwash angle. Vortex is an after effect of this lift. I dont understand how induced drag is reduced by suppressing this vortex.
I would appreciate some clarity on this topic or an explanation which doesn't include vortices.

"...over a finite span." A clue right here, the endplate makes the wing behave as though this condition is less true.

The vortex itself isn't the issue, it's the "leakage" that is resulting in the vortex.  The leakage is from the air near the tip being free to flow around it and results in wasted energy.
The wing is moving along driving air down, but near the tip some of the air, instead of being made to accelerate down, is taking the path of least resistance around the tip (being sucked toward the low pressure side you might say) instead and doesn't contribute to the lift force.  It was displaced, and energy was consumed, but the displacement was out and around the tip instead of down.
The tip plate is just supposed to inhibit some part of the lateral flow around the tip by blocking it, reducing the leakage and resulting in more of the air package being forced downward than without the tip plate.  What is referred to as a reduction in induced drag is just an improvement of the ratio of lift to drag due to less leakage losses, so for a given amount of total lift force, the induced drag is less than for the same lift force without the tip plates.
To do any good, tip plates have to be quite large (because the circulation field extends quite a distance from the wing), which being draggy themselves, cancels out most of the benefit, which is why you don't really see them used very much, except where the end plate effect can be created by default, such as with a T tail where the horizontal stab makes a nice end plate for the top end of the rudder.

Any time you have an object moving through a fluid, you create a disturbance in the fluid (turbulence). The amplitude of disturbance is in proportion to how much energy you are losing (i.e. dissipating into the fluid).
Take a submarine for example. They are designed to optimize the flow of water around the vessel as it moves through the water. This prevents turbulence, which causes noise detectable by Sonar. Creating turbulence, and consequently noise, means that the sub is transferring mechanical energy into it's surroundings. This is wasted energy.
If the vessel (any vessel, boat, airplane, sub) moves faster, typically the turbulence will increase. This means that more energy is being transferred from the sub into the surrounding, and being wasted.
Vortices forming at the wingtips of airplanes is a similar issue. Instead of cleanly "slicing" through the air, the wing is creating turbulence, meaning kinetic energy is being lost to the surrounding air.
Disclaimer: I am not an aerospace engineer.


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What do the colors indicate on this KC135 tanker boom?

What do the colors represent and indicate on the refueling boom?

here's the same picture, caption says "The colorful marking on the boom give the receiver crew a visual indication of the extension of the boom" and that it fluoresces in UV light at night.

That looks like such a bad Photoshop, I thought that's what the answer was going to be!

@tedder42 I know, I was wanting some more information.

They represent the same thing they usually represent anywhere else (at least in Western culture, I am not that familiar with others):

Green = good
Orange = warning
Red = bad

Now, the more interesting question is not what they represent, but what situation could make it turn green, orange, or red?
The part of the boom that is on the right on the photo is the part that is attached to the tanker. The part on the left is the part that attaches to the receiving aircraft.
The left part slides inside of the right part, to give the two aircraft a little bit of room to maneuver. Remember, the boom is rigid, if this telescoping were not possible, the two aircraft would have to stay in exactly the same relative position, up to the millimeter!
So, if the receiving aircraft speeds up and gets closer to the tanker, the inner pipe gets pushed further into the boom, if the receiving aircraft slows down and gets farther away from the tanker, it pulls out the inner pipe from the boom.
The green part is where you normally want the pipe and boom to be. If you get too close to the tanker and risk crashing into it, you push the pipe in further to the orange part and ultimately to the red part. Usually, once you hit the red part, you abort the refueling, and the receiving aircraft breaks off.
Likewise, if you get too far away, you risk ripping off the boom. So, again, as you get further away, you get first to the orange part and then to the red part, at which point, you abort the fueling process.

Getting too far away, you might rip off the boom -- it has happened -- but if the system works as designed, the receptacle on the receiver should disconnect when the extension reaches a limit short of the physical limit.  The boom itself will sense the extension, and sends a signal telling the receptacle to, in essence, "let go of me!"  Works the same, as I recall, as the boom operator commanding a disconnect (which he'd probably also do, to avoid damage to the boom).  Of course, electrons don't always go where they're wanted, so things sometimes go wrong...

"...if the receiving aircraft speeds up and gets closer to the tanker..." Note this isn't always a matter of speed, as such--the boom is at an angle, so (especially if you go through some turbulence) it can deal with up and down motion (particularly common when refueling A-10's, since you have to do so at relatively low altitude).

The boom itself flies, almost like a separate aircraft in itself, hence the built in flight-surfaces on it.  The boom operator is sometimes even referred to as a pilot, and they can move the boom relative to the tanker as needed to try and keep up with the other plane.

The colors correspond with various fore/aft # of feet that define the limits of the extension and compression capacity of the boom. There is also a correlation between the lights on the underside of the KC-135 and the markings on the boom. All are ways of determining the fore/aft position as the receiver pilot during Air-Air Refueling.

I am glad you mentioned the lights on the KC-135. I thought I remembered those from flying the B-52 WST simulator back in the 80s. I did not remember the boom colors, but perhaps it was a limitation of the sim's visual system, or I just forgot.


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Why is destination, flight level, and mach number needed in an oceanic clearance?
When getting an oceanic clearance. The aircraft must announce thier destination airport, their intended flight level, and intended mach cruise speed. Why are these necessary?

Because of the hundreds of other aircraft all in the same airspace?

@757toga This was the original article I read. But after reading deeper I started to understand more. https://skybrary.aero/articles/north-atlantic-operations-atc-clearance#:~:text=There%20are%20three%20elements%20to,or%20following%20a%20random%20route.

Destination, altitude, and speed are essential elements of any flight plan whether VFR, IFR, or Oceanic.
ATC needs this information to ensure proper spacing and sequencing in the case of IFR and Oceanic, and Flight Service needs it for VFR plans in order to know when to launch search and rescue efforts if you are over due landing at your destination.
(Because without these three basic pieces, what else might a flight plan consist of?)

Let's not forget the route of flight. For an aircraft flying through my airspace, that's just as important (or more important if they are landing a long distance away) than the destination.

@RetiredATC, yes, but I wanted to limit my answer to just the elements the OP questioned.


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Why do beechcraft aircraft start with full mixture and throttle?
I've noticed that barons and bonanza's tend to start with full throttle and mixture. Why do they this as opposed to other aircraft that keep throttle at 1/4 inch and put full mixture forward for a few seconds?


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Why does ITT drop drastically after startup?
Is this because of the increase of air that goes through the turbine that cools the parts?

I think you answered yourself

Yes, the increase in airflow has a cooling effect.

Well that was easy!

Yep!  Yes or no questions are tough here because there is a 30 character minimum on answers.  But yeah, your hunch is correct.  Not much more really needs to be said.  (other than maybe thanks for the easy pitch...)


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Why do you have to turn off the prop sync when changing the engine rpm?
I understand how prop sync works, I just don't understand why you have to have to turn it off to switch the rpm. Is there something that I am missing?

Didn't have to do that on the last prop aircraft I flew; can you provide a reference for an aircraft which requires this?

@RalphJ I learned to switch it off in a King Air with a Type I Synchrophaser (a Type II does not require it): "The propellers should be manually synchronized before turning the system on. [...] If you change prop RPM, you may need to switch off the prop sync and repeat this process to resync the props." Nonetheless, it would be great if the OP specified the exact aircraft.

I've answered this previously in regards to why prop sync is required off during takeoff/landing but I suspect this may be different.

Non-FADEC synchrophase systems normally adjust a "slave" propeller, usually No.2 engine, to match the RPM and blade phase of a "master" propeller, usually No. 1 engine.
The only adjusting that is getting done is to the slave propeller's governor, matching its RPM/phase to the master, and there will be a servo unit, typically a torque motor, on that propeller's speed governor that trims the input from that prop's condition lever in the cockpit within a small range.
So when there is a requirement to match RPMs manually before turning synchrophase on, it's because the authority of the slave engine's servo has a limited range. On the Pratt & Whitney PW-120, the setting requirement was only 20 RPM.  If you turn it on with too much differential between left and right propellers, say 50 RPM, the slave propeller's servo may run out of travel trying to match its RPM with the master, and you will run with some RPM differential still in place as the synchrophaser tries to do its thing but can't quite make it.
This will apply to a conventional hydromechanically controlled propeller governor system where the inputs from the cockpit are through linkages to the engine.  If the engine and propeller are controlled through a FADEC system, where the computer has complete control over RPM, you likely won't have this limitation, or at least a much looser one.

So couldn't you just keep it on the whole flight?

Except for having to turn it off when making changes, you could, unless there is a specific limitation, like it has to be off for takeoff.  You could turn it on for climb, and flip it off to come back to cruise, then back on again.  Synchphase is mostly a passenger comfort thing.  The oscillating drone drives ppl nuts.

Sorry for so many questions. would it break if you changed rpm with prop sync on?

No you just won't get them to synch because the servo will just sit at full travel and if that's not enough, the controller will just hold it like that because it's just a dumb box trying to make an internal signal differential go away.


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Why are V speeds the last to be uploaded into a FMC?
When uploading performance information into an FMC, if you put the speeds in before other performance calculations you may get a warning. Why is this?

Because the V speeds are critical to the takeoff performance for the configuration you are departing in, and the FMS won't let you enter them until all the other data is in place, like fuel weights, all up weights, temperature, altitude, etc, etc, etc. It's doing an idiot-proofing function.


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Is fast moving air always low pressure?
Is fast moving air always low pressure? I know this isn't true, but some of the explanations of shock diamonds I see use this analogy.
So how does the speeding up of air over an airfoil create low pressure, but fast moving air doesn't create low pressure by itself? What differs to make fast moving air low pressure? I've also seen that the curve of an airfoil will make low pressure. Why is that? (Picture of shock diamonds below because they look cool)

With a wing, it isn't the moving air that creates low pressure, but the moving airfoil. Air by itself at whatever speed (with no object around to affect its flow) will be constant pressure.

Fluid flow behaves in different ways depending on whether it is subsonic or supersonic.
Subsonic fluid which is not constrained in a constant volume is incompressible, and it's total pressure remains constant. But pressure is a vector, because it is just force per unit area, and the orientation of the surface you are measuring the force against matters, So as you increase velocity, the component of the force per unit area on a surface that is perpendicular to the direction of flow (the dynamic pressure), increases, (the air molecules hit the surface with higher velocity), and the component of the force per unit area on a surface that is parallel to the direction of flow (the static pressure), decreases. The pressure measured by the pitot tube increases as you go faster (indeed that is how airspeed indicators work), and the pressure measured by the static ports decreases (they have to be mounted on a airframe surface that is parallel to the aircraft velocity).
This is the Bernoulli principle. This principle is often stated to say that the faster the flow, the lower the pressure. But when stated this way the pressure it is describing is just the static pressure, not the total pressure. The vector sum of the two components (the total pressure), remains the same. it must remain the same or the principles of conservation of energy and momentum would be violated.
Put a pressure sensor in a flow, and measure the pressure. The direction of the sensor will tell the story. If the sensor points directly into the flow the reading will be higher (it is just a pitot tube). They put these on the nose of the plane as far out in front as possible, to minimize the effect of the airframe itself on the flow hitting the tube. Turn it 90 degrees to the flow. It will now measure static pressure. Point it backwards, the measured pressure will decrease even lower.
Supersonic flow is compressible. The air molecules cannot get "out of the way" of the other molecules that are coming at them at higher than the velocity of the molecules within the fluid due to their heat energy, so they pile up and create a higher density, compressed fluid. Different rules apply.
So, in talking about aerodynamic force on a wing (or on any surface), i.e., Lift, we are talking about the force pushing on the surface, which by necessity, must be pushing perpendicular to the surface, (and to the flow), so, for subsonic velocities, it is the static pressure, which (by Bernoulli) decreases as the flow speed increases.

A little late to the party, but I was going back through this post and re-reading answers and developed a question : On planes, the surrounding air isn't moving relative to the ground, but it is relative to the plane. So the air around a plane shouldn't be low pressure, right? Because there is no pressure field affecting the air around the plane (besides the plane itself) If that doesn't make sense I'll happily clarify.

@Wyatt, ALL physics must be understood as only being applicable, or "true", when examined from within a consistent frame of reference, (FOR). So, yes within the FOR of the ground, the air is not moving and so the pressures do not change. But relative to the aircraft frame the air is moving and the pressures do change. The most basic idea behind Einstein's revolution is that the laws of Physics have to work in ALL frames equivilently. Much of the complexity in doing Physics arises from having to carefully manage how we measure things in the coordinates of the FOR in use.

As an example "thought experiment", consider two scenarios. One plane in flight at a given velocity, and another identical plane, immovably mounted in a giant wind tunnel where the air is moving at that same velocity. Every aerodynamic effect should be the same. The "pressures" you might measure in these two scenarios should be the same, right? YES! Because they depend on what FOR you put the pressure guage in - (how fast is it moving, and its orientation relative to the flow

Oh okay, thanks. So Bernoulli basically states that low pressure air will pull higher pressure air from around it, and in doing so 'stretches' out the air making it low pressure, I think (also speeding it up). I could see that working in a wind tunnel, but how does that work in real flight? The wind tunnel is creating a high pressure at the front, but in real flight there is no such function, right?

To add to that : There's nothing affecting the air besides the plane itself. Now beside the plane, the air isn't getting affected by anything. In a wind tunnel the fan (or whatever wind making device), is making the air low pressure beside the wing, if that makes sense.

Well, no, Bernoulli principle is that Pressure is a VECTOR, and unless constrained by walls or some other boundary, (like supersonic shocks) will remain constant in magnitude. This is required by the conservation of momentum law. Only its components (as measured in different directions), can change. Bernoulli says that as the velocity of the flow increases, the component parallel to the flow, (dynamic or pitot tube pressure), will increase in magnitude, and the component normal (perpendicular) to the flow, (static pressure), will decrease.

To add, you can't apply any physics principle that applies in specific narrow contexts or scenarios, to scenarios in which it does not apply, or requires a more detailed deeper understanding (like "this is about vectors not scalars"), to scenarios where it does not apply without introducing unanticipated errors.

The reason why the principle does not apply to supersonic flow is that the molecules cannot "get out of the way" of the molecules banging into them due to the relative motion of the airfoil, and "bunch up" increasing the density and creating a shock wave. In this shock wave (just like in a closed container), the pressure increases. Since the Pressure is changing, the Ideal gas Law (PV=nRT) applies, and so the temperature goes up, the energy contained within the system is no longer constant, and Bernoulli no longer applies.

so how does speeding up of air over an airfoil create low pressure, but fast moving air doesn't create low pressure by itself?

Last one first.  Fast moving air is the result of air moving from high to low pressure.
Speeding up air over an airfoil creates low pressure by pulling air molecules away from the surface of the wing.  This is because, to a certain extent, air molecules are attracted to one another.  This is known as viscosity.
Air molecules also have mass, and moving them gives momentum, mv.
If one can visit a water fall with a fast flowing river, one can see the momentum of the water carries it past the edge, leaving a gap between the cliff face and the water fall.  In the same way, the momentum of the accelerated air prevents the gap between the airstream and the wing from closing.
In your (excellent) picture, no doubt the exhaust stream will be pulling some of the ambient air along with it.


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Why is spanwise flow more prevalent in the boundary layer?
Spanwise flow is stronger in the boundary layer, but to me it's kind of counterintuitive because the BL flow is slowed down by friction, experiencing less of a pressure gradient over the wing, therefore being decelerated less in region of pressure recovery. Wouldn't this make it so there is less spanwise flow in the boundary layer? (Especially considering that the BL thickens as it passes over the wing)
Answer that might help convey what I am trying to ask.

Does this answer answer your question?

@ROIMaison thanks for your reply. I have seen that answer before, but I still wanted some clarification. (If you click on the link in my question it's actually the same one you provided)

You should draw the velocity triangles on two locations, at the outer edge of the boundary layer and one close to the surface.
Outside of the boundary layer there is a large longitudinal component, so the spanwise component leads to a smaller deflection.
Closer to the surface, the longitudinal velocity is lower, so the same spanwise component leads to a large deflection.

@ROIMaison ah okay, thanks. On this answer, it says that there is an upward speed component from decelerating air on a wing (along with an outward) and that it is more in the BL. Does the same principle apply that you mentioned for that as well?